FINDING THE POINT OF DISCONTINUITY WORKSHEET

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Find any points of discontinuity for each rational function.

Problem 1 :

y = (x + 3)/(x – 4) (x + 3)

Solution

Problem 2 :

y = (x - 2)/(x2 – 4)

Solution

Problem 3 :

y = (x - 3) (x + 1)/(x – 2)

Solution

Problem 4 :

y = 3x(x + 2)/x(x + 2)

Solution

Problem 5 :

y = 2/(x + 1)

Solution

Problem 6 :

y = 4x/(x3 – 9x)

Solution

Problem 7 :

Check the function

g(x) = (x2 + 7x + 10)/(x - 3)(x + 2)

for discontinuities. Conduct appropriate tests to determine if asymptotes exist at the discontinuity values. State the equations of any asymptotes and the domain of g(x)

Solution

Problem 8 :

Match the equation of each rational function with the most appropriate graph. Explain your reasoning.

a) y = (x + 4) / (x2 - 3x - 4)

b) y = (x + 4) / (x2 + 5x + 4)

c) y = (x2 + 4x)/(x + 4)

points-on-discontinuity-q1

Solution

Problem 9 :

Write the equation for each graphed rational function

discontinuity-of-rational-function-q1

Solution

Problem 10 :

Write the equation for each graphed rational function.

discontinuity-of-rational-function-q2.png

Solution

Answer Key

1) x = -3 and 4

2)  x = ±2

3)  x = 2

4)  x = 0, -2

5) x = -1

6)  x = 0, ±3

7) Domain is all real values except -2 and 3.

8) a) Option c

b) Option b

c) Option a

9) y = (x + 2)(x - 3)/(x - 3)

10) y = (x + 5)/(x + 5)(x + 3)

Find the points of discontinuity.  Classify each  point as a vertical asymptote or a hole.

Problem 1 :

y = x/(x2 – 9)

Solution

Problem 2 :

y = (3x2 – 1)/x3

Solution

Problem 3 :

y = (6x2 + 3)/(x – 1)

Solution

Problem 4 :

y = (5x3 – 4)/(x2 + 4x – 5)

Solution

Problem 5 :

y = 7x/(x3 + 1)

Solution

Problem 6 :

y = (12x4 + 10x – 3)/3x4

Solution

Problem 7 :

y = (12x + 24)/(x2 + 2x)

Solution

Problem 8 :

y = (x2 – 1)/(x2 + 3x + 2)

Solution

Problem 9 :

y = (x2 – 1)/(x2 - 2x - 3)

Solution

Answer Key

1)  Vertical asymptotes are at x = 3 and -3, no hole

2)  Vertical asymptotes is at x = 0.No hole

3)  Vertical asymptote is at x = 1.No hole

4)  Vertical asymptotes are at x = -5 and 1.No hole

5)  Asymptote at x = -1.No hole

6)  Asymptote at x = 0. No hole

7)  At x = 0, we have vertical asymptote, hole is at x = -2

8)  vertical asymptote is at x = -2, hole is at x = -1.

9) Vertical asymptote is at x = 3,  hole is at x = -1

State whether or not each of the following functions is continuous.

  • If not, state where the discontinuity occurs and whether or not it is removable.
  • Is the discontinuity is an asymptote, a hole or jump ?
  • If an asymptote, what is its equation ?

Problem 1 :

f(x) = x/(x2 + 1)

Solution

Problem 2 :

f(x) = x/(2x2 - x - 1)

Solution

Problem 3 :

f(x) = (2x + 3)/(x2 - x - 6)

Solution

Problem 4 :

f(x) = (x - 4)/(x2 - 16)

Solution

Problem 5 :

types-of-discontinutyq9.png

Solution

Problem 6 :

types-of-discontinutyq10.png

Solution

Problem 7 :

types-of-discontinutyq11.png

Solution

Problem 8 :

types-of-discontinutyq12.png

Solution

Problem 9 :

Find the value of a if the function is continuous.

types-of-discontinutyq13.png

Solution

Problem 10 :

Find the value of a if the function is continuous.

types-of-discontinutyq15.png

Solution

Problem 12 :

The graph of the function 𝑓(𝑥) is shown below :

types-of-discontinuity-q1

Which of the following statements is true about 𝑓?

I. 𝑓 is undefined at 𝑥 = 1.

II. 𝑓 is defined but not continuous at 𝑥 = 2.

III. 𝑓 is defined and continuous at 𝑥 = 3.

(A) Only I     (B) Only II      (C) I and II     (D) I and III

(E) None of the statements are true.

Solution

Answer Key

1)  

The function is discontinuous at x = -1/2 and 1.

Type of discontinuity = Non removable

Vertical asymptotes are at x = -1/2 and x = 1.

2) 

The function is discontinuous at x = -1/2 and 1.

Type of discontinuity = Non removable

Vertical asymptotes are at x = -1/2 and x = 1.

3)

The function is discontinuous at x = 3 and -2.

Type of discontinuity = Non removable

Vertical asymptotes are at x = 3 and x = -2.

4)  

So, the vertical asymptote is at x = -4

Hole is at x = 4. When x = 4, y = 1/5

5)  The function is not continuous at x = 3, there is removable discontinuity at (3, 6).

6)  Since Lim x->2-  f(x) = Lim x->2+  f(x), then lim x ->2 f(x) does not exists.

Type of discontinuity = Jump discontinuity

7)  

At x = 1, the function is not continuous, then the limit does not exists.

There is jump discontinuity at x = 1.

8)  f(x) is not continuous at x = 3 and x = -3, there is vertical asymptote at x = 3 and x = -3

9)  a = 5

10)  a = 4/3

11)  a = 4 and b = -2

12) Option C

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